The length of a 2 sec pendulum is 1 m.
Given that, initial length of the simple pendulum L₁ = 1 m
Initial time period T₁ = 2 sec
We need to find the length of the pendulum whose time period is 2 sec
T₂ = 2 sec
L₂ = ?
We know that the time period of the simple pendulum is given by the formula,
T = 2π√(L/g)
From the above relation, we can write T ∝ √L
T₁ / T₂ = √(L₁/L₂)
Making L₂ from the above relation, we have,
L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m
Thus, the length of a 2 sec pendulum is 1 m.
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Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = (
× mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s